2016-I-14

$\begin{array}{rcl}
p(2) & = & p(-2) \\
6(2)^4 + 7(2)^3 + a(2)^2 + b(2) +c & = & 6(-2)^4 + 7(-2)^3 +a(-2)^2 +b(-2) +c \\
152 +4a + 2b +c & = & 40+ 4a -2b +c \\
4b & = & -112 \\
b & = & -28
\end{array}$

Also,

$\begin{array}{rcl}
p(x) & \equiv & (lx^2+5x+8)(2x^2+mx+n) \\
& \equiv & 2lx^4 + (10+ml) x^3 + (ln + 5m + 16)x^2 + (5n + 8m) x + 8n \\
\end{array}$

By comparing the coefficients of $p(x)$, we have

$\left\{ \begin{array}{ll}
2l = 6 & \ldots \unicode{x2460} \\
10+ml = 7 & \ldots \unicode{x2461} \\
ln + 5m + 16 = a & \ldots \unicode{x2462} \\
5n + 8m = -28 & \ldots \unicode{x2463} \\
8n=c & \ldots \unicode{x2464}
\end{array} \right.$

From $\unicode{x2460}$, we have $l=3$.

Sub. $l=3$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
10 + m(3) & = & 7 \\
3m & = & -3 \\
m & = & -1
\end{array}$

Sub. $m=-1$ into $\unicode{x2463}$, we have

$\begin{array}{rcl}
5n + 8(-1) & = & -28 \\
5n & = & -20 \\
n & = & -4
\end{array}$

Therefore, $l=3$, $m=-1$ and $n=-4$.

$\begin{array}{rcl}
p(x) & = & 0 \\
(3x^2+5x+8)(2x^2-x-4) & = & 0 \\
\end{array}$

Therefore, $3x^2+5x+8=0$ or $2x^2-x-4=0$.

Consider the discriminant of the equation $3x^2 +5x +8=0$,

$\begin{array}{rcl}
\Delta & = & (5)^2 – 4(3)(8) \\
& = & -71 \\
& < & 0 \end{array}$

Therefore, the equation $3x^2+5x+8=0$ has no real roots.

Consider the discriminant of the equation $2x^2-x-4=0$,

$\begin{array}{rcl}
\Delta & = & (-1)^2 -4(2)(-4) \\
& = & 33 \\
& > & 0
\end{array}$

Therefore, the equation $2x^2-x-4=0$ has two distinct real roots.

Hence, $p(x) = (3x^2+5x+8)(2x^2-x-4)=0$ has two distinct real roots.